When solving real-analysis' problems I like to represent the functions involved and think geometrically what is going on.
Today I got the following exercise :
Let $f \in \mathcal{C}^1(\mathbb{R},\mathbb{R})$, such that : $$\int_\mathbb{R} \mid f \mid \in \mathbb{R}$$ $$\int_\mathbb{R} f'^2 \in \mathbb{R}$$ Proove that $f$ is Hölder continuous.
When I see the assumption involved it's not hard to see that there must be some Cauchy-Schwartz inequality somewhere (because the assumption of absolutely integrable is on the square of the derivative and not only on the derivative itself).
So, with this in mind we easily get the following proof :
We have (using CS) : $$ \mid f(x) -f(y) \mid \leq \int_x^y 1 \times f' \leq \sqrt{\int_x^y f'^2}\sqrt{y-x} $$ Hence it follows that $f$ is $\frac{1}{2}-$Hölder continous since that $\sqrt{\int_x^y f'^2}$ is bounded.
As you noticed it's not hard to come up with the proof since the assumption of the problem on the absolute integrability of the square of the derivative of $f$ directly leads to thinking we must use the Cauchy-Schwartz inequality.
The problem is that I don't like this way of thinking. That's why I am seeking a geometric intuition of the problem.
For example, if we only have the assumption of absolute integrability on $f'$ instead of $f'^2$ does the result still holds?
Why the fact that: $\int_\mathbb{R} f'^2 \in \mathbb{R}$ imply so much regularity for $f$ ?
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Since $f$ is assumed to be $C^1$ the integrability of $(f')^2$ isn't really giving you additional regularity. On any compact interval $[a,b]$ you have by the mean-value theorem $$|f(x) - f(y)| \le |x-y| \sup_{z \in [a,b]} |f'(z)| \le \left( |b-a|^{1/2}\sup_{z \in [a,b]} |f'(z)| \right) |x-y|^{1/2} $$ so that $f$ is locally Holder continuous. The problem is that you lose control of the Holder constant as the interval $[a,b]$ gets large. The integrability condition on $f'$ provides you a way to obtain uniform control over the constant.